A 3.0kg- block sits on top of a 5.0kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force as shown in the figure . The coefficient of static friction between all surfaces is 0.65 and the kinetic coefficient is 0.42.

PART A: What is the minimum value of needed to move the two blocks?

PART B: If the force is 10% greater than your answer for (a), what is the acceleration of each block?

The forces acting on the blocks are:

top block (m1) :
Gravity m1•g (downwards), the normal force on contact between two blocks N1 (upwards), friction force between two blocks F1(fr) ( to the right), Tension T (to the left).
Bottom block (m2).
Gravity m2•g (downwards), gravity m1•g (downwards), the normal force on contact between blocks and the surface N2 (upwards), friction force between two blocks F1(fr) (to the left), friction force between the block and the surface F2(fr) (to the left), tension T (to the left) , unknown force F (to the right).

Equations of the blocks motion (projections on the horizontal and vertical axes):
top block -
x: m1•a = T – F1(fr),
y: 0 = – m1•g + N1. => N1 = m1•g
bottom block –
x: m2•a = F –T –F1(fr) – F2(fr),
y: 0 = - m1•g – m2•g +N2. => N2 = (m1+m2) •g

(a) If the two blocks are just to move, then the force of static friction will be at its maximum, and sothe frictions forces are as follows.
F1(fr) = k1• N1 = k1•m1•g,
F2(fr) = k1• N2 = k1•(m1+m2)•g,

a = 0;
0 = T – F1(fr), => T = F1(fr),
0 = F –T –F1(fr) – F2(fr),
Solve for F.
F = T +F1(fr) + F2(fr) = F1(fr) +F1(fr) + F2(fr) = 2 •F1(fr) + F2(fr) =
= 2• k1•m1•g + k1•(m1+m2)•g = k1•g (3•m1+m2) = 0.65•9.8•(3•3+5) = 89.18 N.

(b) Now force is F1 = 1.1•89.18 = 98.1 N
m1•a = T – F1(fr),
m2•a = F1 –T –F1(fr) – F2(fr),
(m1+m2) •a = F1 - 2 •F1(fr) - F2(fr) =F1 –k2•g•(3•m1+m2),

a = {F1 –k2•g•(3•m1+m2)}/ (m1+m2) =
={98.1 – 0.42•9.8•(3•3+5)}/(3+5) = 5.06 m/s².

To calculate the minimum force needed to move the two blocks, we need to consider the forces acting on the system and determine the maximum static friction force that can be applied.

PART A:

1. Identify the forces acting on the system:
- The weight of the 3.0 kg block (W1 = m1 * g)
- The weight of the 5.0 kg block (W2 = m2 * g)
- The normal force acting on the 5.0 kg block from the surface below (N = W2)

2. Calculate the maximum static friction force (Fs_max):
- Fs_max = coefficient of static friction (μs) * normal force (N)
- Fs_max = 0.65 * N

3. Determine the maximum force needed to overcome static friction:
Since the two blocks are not moving yet, the force to overcome static friction must equal the maximum static friction force (Fs_max). Therefore:
- F_need = Fs_max = 0.65 * N

4. Calculate the normal force (N):
The normal force is equal to the weight of the 5.0 kg block, so:
- N = W2 = m2 * g

5. Substitute the values and calculate the minimum force needed to move the two blocks (F_need):
- F_need = 0.65 * N = 0.65 * (m2 * g)

PART B:

To calculate the acceleration of each block when the force is 10% greater than the minimum force calculated in Part A, we need to use the net force acting on each block and apply Newton's second law.

1. Calculate the net force acting on each block:
- Net force on the 5.0 kg block (F_net2) = force applied (F_applied) - kinetic friction force (F_k2)
- Net force on the 3.0 kg block (F_net1) = kinetic friction force (F_k1)

2. Calculate the kinetic friction forces:
- Kinetic friction force on the 5.0 kg block (F_k2) = coefficient of kinetic friction (μk) * normal force (N)
- Kinetic friction force on the 3.0 kg block (F_k1) = coefficient of kinetic friction (μk) * normal force (N)

3. Calculate the net force acting on each block:
- F_net2 = F_applied - F_k2
- F_net1 = F_k1

4. Apply Newton's second law to calculate the acceleration of each block:
- Acceleration of the 5.0 kg block (a2) = F_net2 / m2
- Acceleration of the 3.0 kg block (a1) = F_net1 / m1

5. Substitute the given values (including 10% greater force) and calculate the accelerations (a1 and a2) using the above formulas.

Remember to convert units if necessary.